To find the kinetic energy of an electron given its de-Broglie wavelength, we use the de-Broglie wavelength formula and the relationship between energy and momentum. The de-Broglie wavelength \(\lambda\) is given by:
\(\lambda = \frac{h}{p}\)
where \(h\) is Planck's constant, and \(p\) is the momentum of the electron.
The momentum \(p\) is related to the kinetic energy \(K\) by:
\(p = \sqrt{2mK}\)
where \(m\) is the mass of the electron.
Substituting this in the de-Broglie formula:
\(\lambda = \frac{h}{\sqrt{2mK}}\)
Squaring both sides, we have:
\(\lambda^2 = \frac{h^2}{2mK}\)
Rearranging for kinetic energy \(K\) gives:
\(K = \frac{h^2}{2m\lambda^2}\)
Let's substitute the known values:
Substitute these values into the equation:
\(K = \frac{(6.63 \times 10^{-34})^2}{2 \times 9.11 \times 10^{-31} \times (2 \times 10^{-10})^2}\)
Calculate the kinetic energy in Joules and then convert it to electron volts:
\(K \approx 6.02 \times 10^{-18} \text{ J}\)
1 eV = \(1.6 \times 10^{-19} \text{ J}\)
So, \(K \approx \frac{6.02 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 37.5 \text{ eV}\)
Hence, the kinetic energy of the electron is 37.5 eV, which confirms that the correct answer is:
37.5 eV